ActualTests.Cisco.640-802.Exam.Q.and.A.08.15.08-DDU (1130589), страница 36
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175.33.3.255D. 26.35.2.255E. 152.135.7.0F. 17.35.36.0Answer: B, D, EExplanation:These are all valid host IP addresses within the /23 subnet.Incorrect Answers:A. This is the network address for the 113.10.4.0/23 subnet.C. This is the broadcast address for the 175.33.2.0/23 subnet.F. This is the network address for the 17.35.36.0/23 subnet.QUESTION 282:The Certkiller network topology is depicted below:Based on the diagram above, which of the following are valid configuration valuesfor the hosts? Select threeA. Host Certkiller A IP address: 192.1.1.85B. Host Certkiller A subnet mask: 255.255.255.224C. Host Certkiller B IP address: 192.1.1.125D. Host Certkiller B default gateway: 192.1.1.85Actualtests.com - The Power of Knowing640-802E.
Host Certkiller C IP address: 192.1.1.166F. Host Certkiller C subnet mask: 255.255.255.224Answer: A, C, FExplanation:The answers A and C are right, because the ip address192.1.1.85 and 192.1.1.125 are in the same subnet192.1.1.64 as the ip address of the subinterface0/1.1.Incorrect Answers:E. This answer is wrong because the network address of the IP address 192.1.1.166 is192.1.1.160.QUESTION 283:Which command on router Certkiller A will assign the last usable IP address fromthe 192.168.32.128/28 subnetwork to a router interface?A.
Certkiller A(config-if)# ip address 192.168.32.142 255.255.255.240B. Certkiller A(config-if)# ip address 192.168.32.143 255.255.255.240C. Certkiller A(config-if)# ip address 192.168.32.158 255.255.255.240D. Certkiller A(config-if)# ip address 192.168.32.145 255.255.255.240E. Certkiller A(config-if)# ip address 192.168.32.144 255.255.255.240F. Certkiller A(config-if)# ip address 192.168.32.158 255.255.255.240G. None of the aboveAnswer: AExplanation:The last usable IP address would be 128 + (16-2) = 142 because only the last 4 bits of thelast octet are used for host addressing.QUESTION 284:The Certkiller LAN is shown below:Actualtests.com - The Power of Knowing640-802A Certkiller .com network administrator is adding host Certkiller 3 to the networkshown in the exhibit. Which IP address can be assigned this host on this network?A.
192.1.1.14B. 192.1.1.18C. 192.1.1.20D. 192.1.1.30E. 192.1.1.31F. 192.1.1.36Answer: B, DExplanation:Subnet Mask of 255.255.255.240 means 4-bits of subnetting. When we do 4-bits ofsubnetting, we have a total of 16 subnets having 16 hosts each. Subnets will be192.1.1.0 ----- 191.1.1.15 (0-15)192.1.1.16 ---- 191.1.1.31 (16-31)192.1.1.32 ---- 191.1.1.47 (32-47)|||||||||192.1.1.240---- 192.1.1.255 (240-255)Only choices B and D are possible as 192.1.1.20 is already used by host Certkiller 1QUESTION 285:A diagram depicting a Certkiller user is shown below:Actualtests.com - The Power of Knowing640-802Based on the information above, which IP address should be assigned to the host?A. 192.168.5.5B.
192.168.5.32C. 192.168.5.40D. 192.168.5.63E. 192.168.5.75F. None of the aboveAnswer: CExplanation:Host address should be in same subnet of Connected Router's Interface. In exhibitRouter's ethernet address is in 192.168.5.33/27 subnet then host address should be also insame subnet.27 bits used for network and 5 bits for host.So Network Address=256-224=32First Subnet 32-64So Host address should be between 32-64 but 32, 64, 63 can't e sued in a host address, asthey are the network and broadcast addresses for the subnet, so only answer C is correct.QUESTION 286:Certkiller is opening a new branch office. Assuming a subnet mask of 255.255.248.0,which three addresses are valid host IP addresses that could be used in this office?(Choose three.)A. 172.16.20.0B.
172.16.24.0C. 172.16.8.0D. 172.16.16.0E. 172.16.31.0Actualtests.com - The Power of Knowing640-802F. 172.16.9.0Answer: A, E, FExplanation:For the 255.255.248.0 subnet mask the following is true.1. 2-2=30 subnets2. 2-2=2,046 hosts per subnet3. 256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.4. Broadcast for the 8.0 subnet is 15.255.
Broadcast for the 16.0 subnet is 23.255, etc.5. The valid hosts are:Subnetfirst hostlast host8.016.024.032.040.048.056.064.08.116.124.132.140.148.156.164.115.25423.25431.25439.25447.25455.25463.25471.254t15.25523.25531.25539.25547.25555.25563.25571.255broadcasReference: http://articles.techrepublic.com.com/5100-6350-5033673.htmlQUESTION 287:A small office Certkiller network is shown below:ipconfig exhibit:Actualtests.com - The Power of Knowing640-802The output shown above is from host CertkillerA.
What value should be displayedfor the Default Gateway of the ipconfig output for this host?A. 172.18.14.6B. 192.168.1.11C. 192.168.1.10D. 192.168.1.254E. 192.168.1.250F. 172.18.14.5G. None of the aboveAnswer: DExplanation:The default gateway setting, which creates the default route in the IP routing table, is acritical part of the configuration of a TCP/IP host. The role of the default gateway is toprovide the next-hop IP address and interface for all destinations that are not located onits subnet. Without a default gateway, communication with remote destination is notpossible, unless additional routes are added to the IP routing table. The default gatewaymust be the router's interface that is on the same IP subnet as the hosts. In this case itrouter Certkiller 2 is the default gateway router, and it's LAN interface with IP address192.168.1.254 would be used.QUESTION 288:The Certkiller network administrator has subnetted the 172.16.0.0 network using asubnet mask of 255.255.255.192.
A duplicate IP address of 172.16.2.121 hasaccidentally been configured on workstation CK1 in this network. The technicianmust assign this workstation a new IP address within that same subnetwork. Whichaddress should be assigned to CK1 ?A. 172.16.1.64B. 172.16.1.80C. 172.16.2.80D. 172.16.2.64E. 172.16.2.127F. 172.16.2.128G. None of the aboveActualtests.com - The Power of Knowing640-802Answer: CExplanation:A subnet mask of 255.255.255.192 (/26) will provide us with 4 subnet (2 usable) eachwith 62 usable hosts per network.
So in our example the four networks will be:172.16.2.1-62172.16.2.65-126172.16.2.129-190172.16.2.193-254Since we know that the host must be in the same IP subnet as 172.16.2.120, only choiceC is correct.QUESTION 289:DRAG DROPCertkiller has three locations and has plans to redesign the network accordingly. Thenetworking team received 192.168.151.0 to use as the addressing for entire networkfrom the administrator. After subnetting the address, the team is ready to assign theaddress.The administrator plans to configure "ip subnet-zero" and use RIP v2 as therouting protocol. As a member of the networking team, you must address thenetwork and at the same time converse unused addresses for future growth.Being mindful of these goals, drag the host addresses on the left to the correct routerinterface.
One of the routers is partially configured. Move the mouse over a routerto view its configuration (** This information is missing**). Not all of the hostaddress choices will be used.Actualtests.com - The Power of Knowing640-802Answer:Actualtests.com - The Power of Knowing640-802QUESTION 290:DRAG DROPCertkiller has three locations and has plans to redesign the network accordingly.
Thenetworking team received 192.168.151.0 to use as the addressing for entire networkfrom the administrator. After subnetting the address, the team is ready to assign theaddress.The administrator plans to configure ip subnet-zero and use RIP v2 as the routingprotocol. As a member of the networking team, you must address the network andat the same time converse unused addresses for future growth.Being mindful of these goals, drag the host addresses on the left to the correct routerinterface. One of the routers is partially configured. Move the mouse over a routerto view its configuration (** This information is missing**).
Not all of the hostaddresses on the left will be used.Actualtests.com - The Power of Knowing640-802Answer:QUESTION 291:If a host on a network has the address 172.16.45.14/30, what is the address of thesubnetwork to which this host belongs?A. 172.16.45.0B. 172.16.45.4C. 172.16.45.8D. 172.16.45.12E. 172.16.45.18Answer: DExplanation:The last octet in binary form is 00001110.
Only 6 bits of this octet belong to the subnetmask. Hence, the subnetwork is 172.16.45.12.QUESTION 292:Which two of the addresses below are available for host addresses on the Certkillersubnet 192.168.15.19/28? (Select two answer choices)A. 192.168.15.17B. 192.168.15.14Actualtests.com - The Power of Knowing640-802C. 192.168.15.29D. 192.168.15.16E. 192.168.15.31F. None of the aboveAnswer: A, CExplanation:The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used forthe networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).The last bit used to make 240 is the 4th bit (16) therefore the first network will be192.168.15.16. The network will have 16 addresses (but remember that the first addressis the network address and the last address is the broadcast address).
In other words, thenetworks will be in increments of 16 beginning at 192.168.15.16/28. The IP address weare given is 192.168.15.19. Therefore the other host addresses must also be on thisnetwork. Valid IP addresses for hosts on this network are: 192.168.15.17-192.168.15.30.Incorrect Answers:B. This is not a valid address for this particular 28 bit subnet mask. The first networkaddress should be 192.168.15.16.D. This is the network address.E. This is the broadcast address for this particular subnet.QUESTION 293:Certkiller has a Class C network and you need ten subnets. You wish to have asmany addresses available for hosts as possible.
Which one of the following subnetmasks should you use?A. 255.255.255.192B. 255.255.255.224C. 255.255.255.240D. 255.255.255.248E. None of the aboveAnswer: CExplanation:Using the 2n-2 formula, we will need to use 4 bits for subnetting, as this will provide for24-2 = 14 subnets. The subnet mask for 4 bits is then 255.255.255.240.Incorrect Answers:A. This will give us only 2 bits for the network mask, which will provide only 2networks.B. This will give us 3 bits for the network mask, which will provide for only 6 networks.Actualtests.com - The Power of Knowing640-802D. This will use 5 bits for the network mask, providing 30 networks.
However, it willprovide for only for 6 host addresses in each network, so C is a better choice.QUESTION 294:You have a single Class C IP address and a point-to-point serial link that you wantto implement VLSM on. Which subnet mask is the most efficient for this point topoint link?A. 255.255.255.0B. 255.255.255.240C. 255.255.255.248D. 255.255.255.252E. 255.255.255.254F. None of the aboveAnswer: DExplanation:For a single point to point link, only 2 IP addresses are required, one for the serialinterface of the router at each end.
Therefore, the 255.255.255.252 subnet mask is oftenused for these types of links, as no IP addresses are wasted.QUESTION 295:You have a network that supports VLSM and you need to reduce IP address wastein your point to point WAN links. Which of the masks below would you use?A. /38B. /30C. /27D. /23E. /18F. /32Answer: BExplanation:For a single point to point link, only 2 IP addresses are required, one for the serialinterface of the router at each end. Therefore, the 255.255.255.252 subnet mask is oftenused for these types of links because no IP addresses are wasted. The subnet mask255.255.255.252 is a /30, so answer B is correct.Incorrect Answers:Actualtests.com - The Power of Knowing640-802A.
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